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An active scalar is a transported scalar field that feeds back on the velocity field, for example a density anomaly producing buoyancy. In an adjoint equations for Boussinesq scalar mixing calculation its dynamical coupling must also be transposed; a passive-scalar sensitivity equation generally omits that feedback.

For a smooth incompressible direct trajectory with total velocity field $\mathbf{U}$ and active scalar $\Theta$, the negative-constraint Lagrangian function in constrained optimization convention givesThe transpose term is the formal adjoint of $\delta \mathbf{u}\cdot \nabla \mathbf{U}$; the two couplings transpose scalar transport by advection and buoyancy. For terminal cost $\int \Theta (T{)}^2$, the terminal data are ${\mathbf{u}}^{†}(T)=0$, ${\theta }^{†}(T)=2\Theta (T)$. They are integrated backward along the stored direct trajectory. No-slip boundary conditions and Dirichlet boundary conditions for the adjoint velocity and homogeneous Neumann boundary conditions for the adjoint scalar remove the spatial boundary terms.

Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.

Baroclinic instability converts potential energy of a stratified sheared basic flow into growing disturbances. In the Eady model, coupled Boundary Rossby waves provide a useful mechanism: their induced velocities displace boundary buoyancy anomalies, and suitable wavelengths permit phase locking. A single neutral boundary wave does not by itself imply exponential instability.

A body force acts throughout a body's or fluid's volume, rather than only through its surface. Gravity is the example used in hydrostatic pressure: a gravitational field $\mathbf{g}$ gives force per unit volume $\rho \mathbf{g}$, where $\rho$ is the mass density. More generally, a body force per unit mass $\mathbf{f}$ contributes $\rho \mathbf{f}$ to momentum conservation. Forces from electromagnetism are another example. Surface pressure forces produce buoyancy on an immersed body, while its body force from gravity acts downward.

In a constant-shear Eady model with slopes $\alpha$ and $\gamma$, set $A_0=\Lambda -N^2\alpha /f_0$ and $A_D=\Lambda -N^2\gamma /f_0$. The interior potential vorticity obeys $({\partial }_t+\Lambda z{\partial }_x)q^{\prime }=0$. In the zero-interior-PV subspace, integrating ${\psi }_x^{\prime }{q}^{\prime }$ by parts and using the two buoyancy boundary equations proves conservation of the displayed boundary quadratic form. Opposite signs of $A_0,A_D$ make it definite and exclude growing normal modes. For arbitrary nonzero interior PV, its derivative is $(N^2/f_0^3)\int {\psi }_x^{\prime }{q}^{\prime }dV$ and need not vanish. The restriction to zero interior PV is automatic for a growing smooth mode with complex phase speed.

With common factors of $\pi$ removed, steady kinematic fluxes satisfy $Q_z=2\alpha \sqrt{M}$, $M_z=FQ/M$ and $F_z=-N^{2}Q$. The Batchelor entrainment hypothesis supplies volume, the integrated buoyancy force supplies momentum, and the environmental mass density gradient reduces buoyancy flux. This uses the Boussinesq approximation.

Boyle's law states that a fixed amount of substance of an ideal gas at constant absolute thermodynamic temperature has constant product of absolute pressure and total volume:Here $p$ is measured relative to a vacuum, not relative to the surrounding atmospheric pressure. Halving the volume doubles the absolute pressure; doubling the volume halves it. A real gas approximately follows Boyle's law when it is sufficiently dilute and far from a phase transition.^[ref]

For an original calculated example, suppose a sealed gas sample has $p_1=100\mathrm{kPa}$ and $V_1=100\mathrm{c}{\mathrm{m}}^3$. After isothermal compression to $V_2=50\mathrm{c}{\mathrm{m}}^3$,Both states have $C=10000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10\mathrm{J}$, since $1\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10^{-3}\mathrm{J}$. The volume ratio is dimensionless, so using the same volume units in numerator and denominator leaves the pressure in kilopascals. Although $pV$ has units of energy, its value is not the compression work: isothermal ideal-gas work involves an integral along the path.

The blue curves show calculated ideal gas states, not measurements. The left panel is a branch of a hyperbola; the right panel plots exactly the same absolute pressures against $1/V$. The dashed line uses gauge pressure with a stipulated constant atmospheric pressure of $100\mathrm{kPa}$. Its displaced intercept explains why absolute pressure in Boyle's law matters.

A short derivation starts from the ideal gas law, $pV=nRT$. Fixing the amount of substance $n$ and absolute thermodynamic temperature $T$ fixes $C=nRT$, where $R$ is the molar gas constant. The microscopic picture in ideal gas pressure from wall collisions is equally appealing: at fixed temperature, the typical particle kinetic energy stays fixed, while compression increases the number density and wall momentum transfer per unit surface area and time. A larger absolute pressure follows without requiring hotter particles.

The constraints are essential. A leak changes the amount of substance. Rapid compression can raise the temperature before heat transfer restores thermal equilibrium, as in the isothermal gas syringe experiment. Molecular size and interactions in a dense real gas invalidate the ideal gas approximation; the Van der Waals equation describes one such correction. Equal endpoint temperatures allow an endpoint test of Boyle's law, but do not establish that every intervening state lies on an isothermal process.

For fixed mass $m$, mass density satisfies $\rho =m/V=mp/C$, so it is proportional to absolute pressure. In a dilute bubbly flow, Boyle's law allows bubble volume and its contribution to buoyancy to increase as hydrostatic pressure decreases, provided gas mass is retained, dissolution is negligible, and the temperature stays approximately fixed. Its usefulness is that a simple ratio predicts a substantial size change without tracking individual molecules.

For a clean inviscid bubble a distance $d\gg a$ below a gravity-restored approximately flat free surface, the surface displacement scales as $F/(\rho g{d}^2)=O(a^3/d^2)$. Zero normal velocity and zero tangential traction are imposed to leading order by an opposite normal image Stokeslet at distance $2d$ from the bubble. Its incident velocity is $-F/(8\pi \mu d)$, which gives the displayed correction at fixed buoyancy. The image can be represented by a same-radius inviscid drop of density $2\rho$ in liquid of density $\rho$. Its deformation-producing strain is $O(F/(\mu d^2))$, so the leading fractional capillary deformation is $O(\mathrm{Bo}(a/d{)}^2)$; small Bond number is the conventional stronger sufficient criterion.

Buoyancy is the resultant upward force exerted by a resting fluid on an immersed body in a downward gravitational field. The fluid pushes inward, perpendicular to each part of the body's surface. Because hydrostatic pressure increases with depth, the upward push from below exceeds the downward push from above. Opposite horizontal forces cancel; buoyancy is the remaining vertical force. A sinking body still receives this upward support. ^[ref]

More precisely, pressure $p$ on a small surface of area $dA$ with outward unit normal vector $\mathbf{n}$ exertsThe minus sign says the force points into the body. On a cube, the top force points down, the bottom force points up, and the side forces cancel in pairs because opposite sides have the same pressure at each depth. This is a surface traction; gravity on the body itself is a separate body force. ^[ref]

For a computed example, stipulate resting water with constant mass density ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$ and uniform gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. A fully immersed cube has side $L=0.10\mathrm{m}$, top depth $h_t=0.20\mathrm{m}$ and bottom depth $h_b=0.30\mathrm{m}$. Each horizontal face has area $A=L^2=0.010{\mathrm{m}}^2$, and its volume is $V=L^3=0.001{\mathrm{m}}^3$. Its top and bottom gauge pressures areThe corresponding forces relative to the common surface pressure are $p_{g,t}A=19.62\mathrm{N}$ downward and $p_{g,b}A=29.43\mathrm{N}$ upward. Therefore the net buoyancy isUsing absolute pressures adds the same $p_{\mathrm{surface}}A$ to both face forces, leaving their difference unchanged. Zero gauge pressure at the surface does not mean zero absolute pressure. Increasing both depths by the same amount also leaves buoyancy unchanged in this constant-mass density model.

Archimedes' principle turns this calculation into a rule for bodies of any shape. It explains floating equilibrium and lets us infer an immersed body's volume from the difference between its force from gravity and the force needed to support it. In particular, buoyancy depends on the displaced volume and the surrounding mass density, whereas the body's own mass density decides how its force from gravity compares with that support.